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Ch 8_handouts - Chapter 8 Applications of Aqueous...

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1 8.1 Solutions of Acids or Bases Containing a Common Ion 8.2 Buffered Solutions 8.3 Exact Treatment of Buffered Solutions 8.4 Buffer Capacity 8.5 Titrations and pH Curves 8.6 Acid-Base Indicators 8.7 Titration of Polyprotic Acids 8.8 Solubility Equilibria and The Solubility Product 8.9 Precipitation and Qualitative Analysis 8.10 Complex Ion Equilibria Chapter 8: Applications of Aqueous Equilibria 2 CHEMISTRY BIOLOGY MEDICINE pH range where buffered solutions play important roles
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3 pH [H 3 O + ] pOH [OH - ] pOH = -log[OH - ] [OH - ] = 10 -pOH pH = -log[H 3 O + ] [H 3 O + ] = 10 -pH K w = 1 x 10 -14 @ 25 °C [H 3 O + ][OH - ] = 1 x 10 -14 pH + pOH = 14 @ 25 °C pH Box 4 A shift in the equilibrium POSITION that occurs because of the addition of an ion already involved in the equilibrium reaction. HNO 2 + H 2 O H 3 O + + NO 2 - Adding NaNO 2 would shift the equilibrium to the left, decreasing the H 3 O + and NO 2 - produced by the reaction between HNO 2 and H 2 O. (Le Châtelier’s principle) Common Ion Effect
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5 Which of the following statements is true? 1. HF dissociates more readily in a solution of NaF than in pure water. 2. HF dissociates equally well in a solution of KCl and in pure water. 3. HF does not dissociate in water to any extent. 6 Nitrous acid, a very weak acid, is only 2.0% ionized in a 0.12 M solution. Calculate the [H + ], the pH, and the percent dissociation of HNO 2 in a 1.0 M solution that is also 1.0 M in NaNO 2 ! HNO 2(aq) H + (aq) + NO 2 - (aq) K a = = 4.0 x 10 -4 [H + ] [NO 2 - ] [HNO 2 ] Initial Concentration (mol/L) Equilibrium Concentration (mol/L) [HNO 2 ] 0 = 1.0 M [HNO 2 ] = 1.0 – x (from dissolved HNO 2 ) [NO 2 - ] 0 = 1.0 M [NO 2 - ] = 1.0 + x (from dissolved NaNO 2 ) [H + ] 0 0 [H + ] = x (neglect the contribution from water) Like Example 8.1 - I
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7 K a = = = 4.0 x 10 -4 [H + ] [NO 2 - ] [HNO 2 ] ( x ) ( 1.0 + x ) (1.0 – x ) Assume (1.0 + x) 1.0 and (1.0 – x) 1.0 : x (1.0) (1.0) = 4.0 x 10 -4 or x = 4.0 x 10 -4 = [H + ] Therefore pH = - log [H + ] = - log (4.0 x 10 -4 ) = 3.40 The percent dissociation is: 4.0 x 10 -4 1.0 x 100 = 0.040 % Nitrous acid Nitrous acid alone + NaNO 2 [H + ] 2.0 x 10 -2 4.0 x 10 -4 pH 1.70 3.40 % Diss 2.0 0.040 Like Example 8.1 - II 8 A buffer is a solution that contains concentrations of both components of a conjugate acid-base pair . When small quantities of H + or OH - are added to the buffer, they cause a small amount of one buffer component to convert into the other (HA A - or A - HA). Buffers resist a change in pH . As long as the amounts of H 3 O + and OH - are small (compared to the concentrations of the acid and base in the buffer), the added ions will have little effect on the pH since they are consumed by the buffer components . The ability of a buffer to resist a change of pH is due to the buffer capacity . pH equals pKa IF the acid-to-base ratio in the buffer is 1:1 What is a BUFFER?
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9 10 Modified pH Step 1: Do stoichiometric calculations to determine new concentrations. Assume reaction with H + /OH - goes to completion. Original buffered solution pH Step 2: Do equilibrium calculations.
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